Calculate Geometric Trial Probabilities

Enter success probability (p) and trial number of 1st success (k).

Probability p between 0 and 1 (e.g. 0.2 = 20%).
Trial $k$ ($k \ge 1$).

Calculation Results

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Mathematical Standard--

Calculated using geometric distribution probability mass function: P(X = k) = (1 - p)^{k - 1} \cdot p

*Note: The geometric distribution models the number of Bernoulli trials required until the first success occurs (memoryless property).

Quick Summary

The Geometric Distribution Calculator evaluates exact probability $P(X = k)$, cumulative probabilities $P(X \le k)$, expected trials to first success ($1/p$), and variance.

Formula Explanation

P(X = k) = (1 - p)^{k - 1} \cdot p
P(X \le k) = 1 - (1 - p)^k
E(X) = \frac{1}{p}, \quad \text{Var}(X) = \frac{1 - p}{p^2}

How It Works

The geometric distribution models the number of trials needed to achieve the first success in repeated independent Bernoulli trials. The Geometric Distribution Calculator computes exact probability $P(X=k)$, cumulative distribution $P(X \le k)$, and expected waiting trials $E(X) = 1/p$.

Step-by-Step Worked Example

Practical Problem: A basketball player makes free throws with $p = 0.20$ (20% success rate). Calculate probability that the first successful shot occurs on trial $k = 3$.

  1. Step 1: Calculate failure probability ($1 - p$): $1 - 0.20 = \mathbf{0.80}$.
  2. Step 2: Calculate failure factor for first $k - 1 = 2$ trials: $0.80^2 = \mathbf{0.64}$.
  3. Step 3: Multiply by success probability on 3rd trial ($p = 0.20$): $P(X = 3) = 0.64 \times 0.20 = \mathbf{0.1280\text{ (12.80\%)}}.$
  4. Step 4: Calculate Cumulative $P(X \le 3)$: $1 - 0.80^3 = 1 - 0.512 = \mathbf{0.4880\text{ (48.80\%)}}.$
  5. Step 5: Calculate Expected Trials $E(X)$: $E(X) = 1 / 0.20 = \mathbf{5.00\text{ trials required on average}}$.

Real-World Calculation Examples

Scenario 1: Basketball Free Throws (p = 0.20, k = 3)

Parameters: p = 0.20, k = 3
Result: P(X = 3) = 0.1280 (Expected trials E(X) = 5.00).

Scenario 2: Rolling a 6 on a Die (p = 1/6)

Parameters: p = 0.1667, k = 1
Result: P(X = 1) = 0.1667 (Expected trials E(X) = 6.00).

Scenario 3: Machine Component Failure Inspection

Parameters: p = 0.05 defect rate, k ≤ 10
Result: P(X ≤ 10) = 0.4013.

Scenario 4: Sales Prospecting First Conversion

Parameters: p = 0.10 conversion, k = 1st call
Result: Expected calls E(X) = 10 calls.

Key Benefits of Using This Calculator

Expected Waiting Trials ($1/p$)

Calculates average expected trials needed until first success.

Survival & Cumulative Probabilities

Outputs $P(X \le k)$ and survival probability $P(X > k) = (1-p)^k$.

Memoryless Property Modeling

Accurately models memoryless discrete trial sequences.

100% Free & Client-Side

Executes locally in your browser with zero latency or web server transmission.

Frequently Asked Questions (FAQ)

What is a geometric distribution?

The geometric distribution models the probability of needing $k$ independent Bernoulli trials to get the first success.

What is the memoryless property of geometric distribution?

The memoryless property means past unsuccessful trials do not affect future trial probabilities ($P(X > s + t \mid X > s) = P(X > t)$).

What is formula for expected value and variance?

$E(X) = \frac{1}{p}$ and $\text{Var}(X) = \frac{1 - p}{p^2}$.

How do I calculate geometric probabilities in Excel?

Use formula =NEGBINOM.DIST(k - 1, 1, p, FALSE) for exact $P(X=k)$.

What is difference between geometric and negative binomial distribution?

Geometric distribution models trials for the 1st success; negative binomial models trials for $r$-th success ($r \ge 1$).

What is continuous equivalent of geometric distribution?

The Exponential distribution is the continuous memoryless counterpart of the discrete geometric distribution.

Can trial k be 0?

In this formulation (number of trials), $k \ge 1$. In alternative formulation (number of failures before 1st success), $k \ge 0$.

What is survival function $P(X > k)$?

Survival function $P(X > k) = (1 - p)^k$ is the probability that the first $k$ trials all fail.

What is median of a geometric distribution?

$\text{Median} = \left\lceil \frac{-\ln 2}{\ln(1 - p)} \right\rceil$.

Why is geometric mean different from geometric distribution?

Geometric mean is a central tendency average ($\sqrt[n]{\prod x_i}$); geometric distribution is a discrete probability distribution for first success trials.